Is $[F(x)]^{-1}=F(-x)$ obvious for the given matrix $F(x)$? 
$$F(x)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}$$

Is it very obvious (in the sense that without any calculations) that $[F(x)]^{-1}=F(-x)$? 
My book directly writes this without any explanation. How is this evident without calculation?
 A: Yes, it is obvious if you know multiplication by blocks and the matrix of a rotation (of angle $x$) in the plane:
$$R(x)=\begin{bmatrix}\cos x&-\sin x\\\sin x & \cos x\end{bmatrix}, \qquad R(x)^{-1}=R(-x)=\begin{bmatrix}\cos x&\sin x\\-\sin x & \cos x\end{bmatrix}.$$
Now, consider your matrix as a block matrix:
$$F(x)=\begin{bmatrix}R(x)&\mathbf 0 \\
\mathbf 0 & 1\end{bmatrix},$$
where $\mathbf 0$ denotes the null $2{\times} 1$ column vector in the first row and the null $1{\times}2$ row vector in the second row . Using block multiplication, we see at once that this matrix is invertible and its inverse is
$$F(x)^{-1}=\begin{bmatrix}R(x)^{-1}&\mathbf 0 \\
\mathbf 0 & 1\end{bmatrix}=\begin{bmatrix}R(-x)&\mathbf 0 \\
\mathbf 0 & 1\end{bmatrix}. $$
A: This matrix $F(\theta)$ represents a linear transformation that fixes the $z$ axis and rotates the $xy$-plane by an angle of $\theta$ around the origin. The inverse mapping is of course therefore fixing $z$ and rotating the $xy$-plane by an angle of $-\theta$, which is precisely $F(-\theta)$!
